A $3\times3$ matrix $P$ is such that, $P^3 = P$. Then the eigenvalues of $P$ are? Recently I came across a question here p.25 question 7. This is a question from the GATE EE 2016 exam. But I was having difficulty in analyzing the question as to whether it has a definite answer or not based on the given information in the question. Below I share two approaches of my peers and two of them came up with two different approaches with two different answers.


A $3\times3$ matrix $P$ is such that, $P^3 = P$. Then the eigenvalues of $P$ are:
$\quad A. \quad1, 1, -1$
$\quad B.\quad1, 0.5 + j0.866, 0.5 - j0.866$
$\quad C \quad1, -0.5 + j0.866, -0.5 - j0.866$
$\quad D \quad 0, 1, -1$
$\tag {$\text{GATE 2016 EE Set 2}$}$

This is a single option correct question.
One of my peers came up with the following approach:

Let $\lambda$ be a eigen value of matrix $P$ and $X$ be the corresponding eigen vector so:
$$PX=\lambda X \tag 1$$
$$\implies P^3 X=\lambda^3 X$$
$$\implies PX=\lambda^3X \quad\text{[As it is given that $P^3=P$]}\tag 2$$
From $(1)$ and $(2)$ we have:
$$\lambda^3X=\lambda X$$
$$\implies (\lambda^3-\lambda)X=\mathbb{O}_{3\times 1}$$
But as $X \neq \mathbb{O}_{3\times 1}$ we have:
$$\lambda^3-\lambda=0 \implies \lambda=0,1,-1$$
So option (D) seems to be correct

But my another peer came up with a specific example:

Let $P=\begin{pmatrix}
1 & 0 & 0\\
0 & 0 & 1 \\
0 & 1 & 0
\end{pmatrix}$,
so $P^2=\mathbb{I}_{3\times 3}$ $\implies P^3=P$
Let $\lambda$ be a eigen value of matrix $P$ and $X$ be the corresponding eigen vector so:
$$PX=\lambda X $$
$$\implies (P-\lambda\mathbb{I}_{3\times 3})= X$$
For $X$ to non zero we should have:
$$det(P-\lambda\mathbb{I}_{3\times 3})=0$$
$$\implies\begin{vmatrix}
1-\lambda & 0 & 0\\
0 & 0-\lambda & 1 \\
0 & 1 & 0-\lambda
\end{vmatrix}=0$$
$$\implies \lambda=-1,1,1$$


What is the problem which has crept in the solution? I guess I and my peers are missing out something. What are we missing out so that we are unable to generate all possible viable solutions?
I myself tried to approach as follows:

$$P^3=P$$ $$\implies |P^3|=|P|$$ $$\implies |P|^3=|P|$$$$\implies |P|^3-|P|=0$$ $$\implies |P|=0,1,-1$$
For case: $|P|=0$ I have one eigen value as $0$,but I have clue as to how to find the other two.
For the other two cases of $|P|$ I could not proceed further...

 A: The solution by your first peer tells you that eigenvalues can only be $0$, $1$ or $-1$.
But this doesn't mean they are exactly that, in fact any combination $(\lambda_1,\lambda_2,\lambda_3)$ with $\lambda_i\in\{-1,0,1\}$ is possible.
With $(0,0,0)$ a possible candidate is the zero matrix , with $(1,1,1)$ a possible candidate is the identity matrix, with $(-1,1,1)$ you second peer example, etc.
So we can rule out answers $B$ and $C$ for sure.
Without any more precisions $A$ or $D$ could fit, but as said previously it's not even mandatory.
Adding the reasonable condition $P^2\neq P$ would rule out the zero and identity matrix, but it still doesn't make a difference in regard to choices $A, D$ has both $\operatorname{diag}(1,1,-1)$ and $\operatorname{diag}(0,1,-1)$ would satisfy the problem.
So the question is definitely ill-posed.

Edit: note for your solution since eigenvalues are among $\{-1,0,1\}$ the product of three of such integers is among the same set, and so is the determinant. But this does not really help you in the reciprocal problem of finding the eigenvalues knowing only the possible values of the determinant.
In fact the approach given by your first peer is I think the best one, you just misinterpreted the result considering the eigenvalues should all be distinct, while you just need to select them in this $\{-1,0,1\}$ pool with possible repetitions.
A: If $P$ is not the zero matrix, and as zwim noted, if $P^2 \ne P$ and $P^3=P,$ then we can use the Cayley-Hamilton theorem, since $P^3-P=0_{n\times n}$ this means that your eigenvalues would be the roots of $\lambda^3-\lambda = 0.$ But there a lot of ifs and I agree that the problem is not properly stated.
A: You have,
$$
P^3=P\\
P^{-1}P=1\\
$$
where $P^{-1}$ is the inverse of $P$ and the $1$ in the equation represents a 3x3 matrix with ones in the diagonal and zeros in off diagonal elements (the assumtion is that $P$ is invertible).
$$
P^{-1}P^3=P^2=P^{-1}P=1\\
P^{-1}P^2=P=P^{-1}\\
P=P^{-1}
$$
That is $P$ is "involutory".  Write the eigenvalue equation,
$$
Px=\lambda x\\
P^{-1}Px=\lambda P^{-1}x\\
x=\lambda^2 x \\
(1-\lambda^2)x=0
$$
and thus the eigenvalues can only be $\pm 1$
